A cell of electromotive force (e.m.f.) E and negligible internal resistance is connected into a circuit, as shown.

Question 18

A cell of electromotive force (e.m.f.) E and negligible internal resistance is connected into a circuit, as shown.

The voltmeter has a very high resistance and reads a potential difference Vout.

What is the ratio Vout / E ?

A 1 / 6             B 1 / 3             C 1 / 2             D 2 / 3

Reference: Past Exam Paper – March 2018 Paper 12 Q37

Solution:

Answer: D.

The voltmeter reads the p.d. across the 4 Ω resistor.

We can identify 2 loops in the circuit.

Loop 1: Cell - 12 Ω resistor

Loop 2: Cell - 2 Ω resistor – 4 Ω resistor

In the second loop, the 2 Ω and 4 Ω resistors are in series to each other.

From Kirchhoff’s law, the sum of p.d. in each loop is equal to the e.m.f. in the circuit.

The p.d. V_out across the 4 Ω resistor can be obtained by the potential divider equation.

V_out = [4 / (4+2)] × E = 4E / 6 = 2E / 3

 

Ratio = V_out / E = (2E/3) / E = 2 / 3